Ncert Solutions Physics Class 12th
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6.10 A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°.
6.10 A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°.
6.10 Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span, l = 25 m
Earth’s magnetic field, B = 5 T
Dip angle, = 30
The vertical component of the earth’s magnetic field, = B = 5 = 2.5 T
Voltage difference between the ends of the wing
6.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
6.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
6.9 Mutual inductance of a pair of coils, = 1.5 H
Initial current, = 0 A
Final current, = 20 A
Change in current, dI = - = 20 A
Time taken for change, dt = 0.5 s
From the relation of induced emf e = , where is the change in the
6.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
6.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
6.8 Given:
Initial current, = 5.0 A
Final current, = 0 A
Change in current, di = = 5 A
Time taken for the change, dt = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, L is given by
L = = = 4 H
Hence, the self-inductance of the coil i
What are integrated circuits (ICs), and what are their advantages?
What are integrated circuits (ICs), and what are their advantages?
As per the NCERT Textbooks
“An integrated Chip consists of many passive and active components fabricated on a single chip of silicon. these ICs are compact, low-cost, and highly reliable. They consume less power and have high speed.”
These ICs are used in majority of our daily use electronics and
2.36 Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm–1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm–1 at its surface in the downward direction, corresponding to a surface charge density = –10–9 C m–2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
2.36 Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm–1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?
(Hint: The earth has an electric field of about 100 Vm–1 at its surface in the downward direction, corresponding to a surface charge density = –10–9 C m–2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
2.36 We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Yes, the person will get an electric shock if he touches the metal slab next morning. The steady discharging current
2.35 A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
2.35 A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge . For positive charge , potential difference V
2.34 Describe schematically the equipotential surfaces corresponding to
(a) A constant electric field in the z-direction,
(b) A field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) A single positive charge at the origin, and
(d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.
2.34 Describe schematically the equipotential surfaces corresponding to
(a) A constant electric field in the z-direction,
(b) A field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) A single positive charge at the origin, and
(d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.
2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.
Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases
Concentric spheres centered at the origin are equipotential surfaces.
A periodically
2.33 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm–1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should
like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
2.33 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm–1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should
like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
2.33 Potential rating of the capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, = 3
Dielectric strength = V/m
For safety, the field intensity should not cross 10% of the dielectric strength, hence
Electric field intensity, E = 10 % of = V/m
Capacit
2.32 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
2.32 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
2.32 Length of the co-axial cylinder, l = 15 cm = 0.15 m
Radius of the outer cylinder, = 1.5 cm = 0.015 m
Radius of the inner cylinder, = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 C
Capacitance of a co-axial cylinder of radii and
Why is a p-type semiconductor electrically neutral despite having more holes than electrons?
Why is a p-type semiconductor electrically neutral despite having more holes than electrons?
A p-type semiconductor is electrically neutral despite having more holes, because the number of positively charged holes is exactly balanced/equal by the number of negatively charged acceptor ions introduced during doping. so practically untill any volatage is applied the semiconductor remains